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Re: Gravitational redshift and clocks



On Mon, 1 Sep 2003, John S. Denker wrote:

On Mon, 1 Sep 2003, Savinainen Antti wrote:
>>General Relativity predicts that when a light ray leaves the
>>ground and rises higher its frequency gets smaller (i.e., light
>>is redshifted).

OK.

On 09/01/2003 03:24 PM, Stephen Speicher wrote:
> No. General relativity does not predict a gravitational effect on
> the frequency itself, i.e., the frequency does not itself "get[s]
> smaller." In general relativity the frequency is not intrinsic to
> the object -- it does not itself get smaller or get larger -- but
> rather it is a _relationship_ between observer and object. The
> observed redshift is a consequence of the nonlocal path through
> the curvature of the spacetime manifold, not due to any change in
> the local frequency itself.

That's mostly wrong and entirely irrelevant.

There is a perfectly well-defined notion of local
frequency.

The curvature of spacetime is not observable locally; it is only
over nonlocal regions of spacetime where the effects of
gravitation are observable.

One needs to specify who's measuring what frequency in which
reference frame, but that's just routine housekeeping.


Here you repeat, in your own words, essentially what I said
above, except you do not draw the proper inferences. In general
relativity frequency, just like energy, is not intrinsic to an
object, but rather is a relationship between the observer and the
object. The frequency, or energy, of a photon does not itself
'get smaller' or 'get larger': there is no 'thing' called
frequency or energy which itself changes -- they are both
relationships dependent on the observer. An intrinsic property
does not depend on how it is observed. Energy and frequency are
not invariant, but there is invariant mass.

To first order (which is all that matters here)
the redshift depends on the potential difference
phi(X) - phi(Y)
with no other dependence on the "nonlocal path
through the curvature of the spacetime manifold"
that the photon took getting from X to Y.

Specifically,
delta(lambda) / lambda = delta(phi)


General relativity is not Newtonian mechanics; in GR there is no
well-defined notion of gravitational potential -- one simply
cannot define a tensorial local gravitational potential. Relating
\phi to g_00 requires both an approximation as well as
specifically-crafted coordinates. There is no gravitational
potential \phi in the field equations of general relativity as
there is a Newtonian gravitational potential in Newtonian
mechanics.

Note: I am well-aware of older texts on this subject -- ones
which also use out-moded terms such as gravitational 'force' --
but the modern notions are much more firmly grounded. In modern
general relativity there is no meaningful notion of energy
density of the gravitational field. See a modern text, such as
Wald's "General Relativity."

--
Stephen
speicher@caltech.edu

Ignorance is just a placeholder for knowledge.

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